CAIE M1 2024 November — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeEnergy method - smooth inclined plane (no resistance)
DifficultyModerate -0.8 Part (a) is a direct application of conservation of energy (PE = KE) requiring only one equation and basic algebra. Part (b) adds friction using the work-energy theorem but remains a straightforward calculation with given values. Both parts are standard textbook exercises with no problem-solving insight required, making this easier than average for A-level.
Spec6.02i Conservation of energy: mechanical energy principle
\includegraphics{figure_2} A particle of mass \(7.5\) kg, starting from rest at \(A\), slides down an inclined plane \(AB\). The point \(B\) is \(12.5\) metres vertically below the level of \(A\), as shown in the diagram.
  1. Given that the plane is smooth, use an energy method to find the speed of the particle at \(B\). [2]
  2. It is given instead that the plane is rough and the particle reaches \(B\) with a speed of \(8 \text{ ms}^{-1}\). The plane is \(25\) m long and the constant frictional force has magnitude \(F\) N. Find the value of \(F\). [3]
Question 2:
AnswerMarks
2(a)1
KE = 7.5v2
2
AnswerMarks Guidance
PE =7.5g12.5 [= 937.5]*B1 Either correct.
v=15.8ms−1DB1 5 10.
12.5
SC B1 for v2 =02 +2(gsin)  v=15.8 (or with cos).
sin
B0 for v2 =02 +2g12.5v=15.8 or correct answer with no
working.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
2(b)KE =0.57.582 [= 240]
BB1
7.5g12.5=0.57.582 +F25M1 Attempt at work energy equation; 3 terms; dimensionally correct;
allow sign errors.
AnswerMarks
F =27.9A1
ALTERNATIVE FOR 2(b)
AnswerMarks Guidance
82 =02 +2a25a=1.28B1 Finding the correct acceleration down the plane.
12.5
7.5g −F =7.5a
AnswerMarks Guidance
25M1 Newton’s second law parallel to the plane; allow sign errors and
sin/cos mix on the weight component; dimensionally correct.
Allow with their a, or just a .
AnswerMarks
F =27.9A1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | 1
KE = 7.5v2
2
PE =7.5g12.5 [= 937.5] | *B1 | Either correct.
v=15.8ms−1 | DB1 | 5 10.
12.5
SC B1 for v2 =02 +2(gsin)  v=15.8 (or with cos).
sin
B0 for v2 =02 +2g12.5v=15.8 or correct answer with no
working.
2
Question | Answer | Marks | Guidance
--- 2(b) ---
2(b) | KE =0.57.582 [= 240]
B | B1
7.5g12.5=0.57.582 +F25 | M1 | Attempt at work energy equation; 3 terms; dimensionally correct;
allow sign errors.
F =27.9 | A1
ALTERNATIVE FOR 2(b)
82 =02 +2a25a=1.28 | B1 | Finding the correct acceleration down the plane.
12.5
7.5g −F =7.5a
25 | M1 | Newton’s second law parallel to the plane; allow sign errors and
sin/cos mix on the weight component; dimensionally correct.
Allow with their a, or just a .
F =27.9 | A1
3
Question | Answer | Marks | Guidance
\includegraphics{figure_2}

A particle of mass $7.5$ kg, starting from rest at $A$, slides down an inclined plane $AB$. The point $B$ is $12.5$ metres vertically below the level of $A$, as shown in the diagram.

\begin{enumerate}[label=(\alph*)]
\item Given that the plane is smooth, use an energy method to find the speed of the particle at $B$. [2]

\item It is given instead that the plane is rough and the particle reaches $B$ with a speed of $8 \text{ ms}^{-1}$. The plane is $25$ m long and the constant frictional force has magnitude $F$ N.

Find the value of $F$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q2 [5]}}
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